Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
vector<int> add(vector<int> &a, vector<int> &b) {
  vector<int> c;
  int i = 0, j = 0;
  for (i = 0, j = 0; i < a.size() && j < b.size();) {
    if (a[i] <= b[j])
      c.push_back(a[i++]);
    else
      c.push_back(b[j++]);
  }
  while (i < a.size()) c.push_back(a[i++]);
  while (j < b.size()) c.push_back(b[j++]);
  while (c.size() > 10) c.pop_back();
  return c;
}
struct RMQ {
  vector<vector<int> > t;
  int n;
  RMQ(int n) : n(n) { t.resize(n << 1); }
  void modify(int p, int v) {
    if (t[p + n].size() > 10) return;
    for (t[p += n].push_back(v); p > 1; p >>= 1)
      t[p >> 1] = add(t[p], t[p ^ 1]);
  }
  vector<int> query(int l, int r) {
    vector<int> res;
    for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
      if (l & 1) res = add(res, t[l++]);
      if (r & 1) res = add(res, t[--r]);
    }
    return res;
  }
} S(N);
int par[N], sz[N], lvl[N], idx[N], ridx[N], head[N], tme;
vector<int> adj[N], A[N];
void dfs(int u, int p) {
  int mx = 0;
  sz[u] = 1;
  par[u] = p;
  lvl[u] = lvl[p] + 1;
  for (int i = 0; i < adj[u].size(); i++)
    if (adj[u][i] != p) {
      int v = adj[u][i];
      dfs(v, u);
      sz[u] += sz[v];
      if (sz[v] > mx) swap(adj[u][0], adj[u][i]), mx = sz[v];
    }
}
void hld(int u, int p) {
  if (adj[u].size() && adj[u][0] != p) {
    int c = adj[u][0];
    idx[c] = ++tme;
    ridx[tme] = c;
    head[c] = head[u];
    hld(c, u);
  }
  for (int i = 1; i < adj[u].size(); i++)
    if (adj[u][i] != p) {
      int v = adj[u][i];
      idx[v] = ++tme;
      ridx[tme] = v;
      head[v] = v;
      hld(v, u);
    }
}
vector<int> query(int u, int v) {
  vector<int> ans;
  while (true) {
    if (head[u] == head[v]) {
      if (lvl[v] < lvl[u]) swap(u, v);
      vector<int> k = S.query(idx[u], idx[v]);
      return add(ans, k);
    }
    if (lvl[head[u]] < lvl[head[v]]) swap(u, v);
    vector<int> k = S.query(idx[head[u]], idx[u]);
    ans = add(ans, k);
    u = par[head[u]];
  }
}
int main() {
  int n, m, q;
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  head[1] = 1;
  idx[++tme] = 1;
  hld(1, 0);
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    S.modify(idx[x], i);
  }
  vector<tuple<int, int, int, int> > Q;
  for (int i = 1; i <= q; i++) {
    int u, v, a;
    scanf("%d %d %d", &u, &v, &a);
    if (v < u) swap(u, v);
    Q.push_back(make_tuple(u, v, a, i));
  }
  sort(Q.begin(), Q.end());
  vector<int> xx;
  for (int i = 0; i < Q.size(); i++) {
    int u, v, a, idx;
    tie(u, v, a, idx) = Q[i];
    if (i > 0 && get<0>(Q[i - 1]) == get<0>(Q[i]) &&
        get<1>(Q[i - 1]) == get<1>(Q[i])) {
      A[idx] = xx;
      A[idx].resize(min((int)xx.size(), a));
    } else {
      xx = query(u, v);
      A[idx] = xx;
      A[idx].resize(min((int)xx.size(), a));
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%d ", (int)A[i].size());
    for (int x : A[i]) printf("%d ", x);
    printf("\n");
  }
}